Respuesta :
N = number of nickles
D = number of dimes = N+4
Q = number of quarters
= 25-N-D
= 25-N-(N+4)
= 21-(2N)
Add the coins to get $3.20
(N*$0.05) + (D*$0.10) + (Q*$0.25) = $3.20
Expressing D and Q in terms of N:
(N*$0.05) + ((N+4)*$0.10) + (21-2N)*$0.25=$3.20
Solve for N. Then plug N into the equations for D and Q
D = number of dimes = N+4
Q = number of quarters
= 25-N-D
= 25-N-(N+4)
= 21-(2N)
Add the coins to get $3.20
(N*$0.05) + (D*$0.10) + (Q*$0.25) = $3.20
Expressing D and Q in terms of N:
(N*$0.05) + ((N+4)*$0.10) + (21-2N)*$0.25=$3.20
Solve for N. Then plug N into the equations for D and Q
Using a system of equations, it is found that she had 7 nickels, 11 dimes and 7 quarters.
What is a system of equations?
A system of equations is when two or more variables are related, and equations are built to find the values of each variable.
In this problem, the variables are:
- Variable x: Number of nickels.
- Variable y: Number of dimes.
- Variable z: Number of quarters.
In total, she had 25 coins, hence:
[tex]x + y + z = 25[/tex]
The coins were worth $3.20, hence:
[tex]0.05x + 0.1y + 0.25z = 3.2[/tex]
The number of dimes exceeded the number of nickels by 4, hence:
[tex]y = x + 4[/tex]
Then:
[tex]x + y + z = 25[/tex]
[tex]x + x + 4 + z = 25[/tex]
[tex]2x + z = 21[/tex]
[tex]z = 21 - 2x[/tex]
Replacing on the second equation, we can find x:
[tex]0.05x + 0.1y + 0.25z = 3.2[/tex]
[tex]0.05x + 0.1(x + 4) + 0.25(21 - 2x) = 3.2[/tex]
[tex]-0.35x = -2.45[/tex]
[tex]0.35x = 2.45[/tex]
[tex]x = \frac{2.45}{0.35}[/tex]
[tex]x = 7[/tex]
Then, for y and z:
[tex]y = x + 4 = 7 + 4 = 11[/tex]
[tex]z = 21 - 2x = 21 - 2(7) = 7[/tex]
Hence, she had 7 nickels, 11 dimes and 7 quarters.
To learn more about a system of equations, you can take a look at https://brainly.com/question/14183076
